How should I understand this explanation that $\mathbb{Q}$ and $\mathbb{N}$ aren't homeomorphic subspaces of $\mathbb{R}$

Question

The definition of open sets I'm working with (from the Napkin) is:

A set $U \subseteq M $ is open if for all $p \in U$, some $r$-neighborhood ($r > 0$) exists in $U$.

The question I'm trying to solve is this:

Are $\mathbb{Q}$ and $\mathbb{N}$ homeomorphic subspaces in $\mathbb{R}$?

My response was no, because any subset of $\mathbb{N}$ is closed, since any sequence that converges using points in $\mathbb{N}$ converges to something in $\mathbb{N}$, whereas there are subsets of $\mathbb{Q}$ that aren't closed - i.e. $3, 3.1, 3.14, 3.141, 3.1415 ... $ that converge to $\pi \in \mathbb{Q'}$.

But the Napkin's solution I don't understand - it simply says that no singleton set is open in $\mathbb{Q}$, but all singleton sets are open in $\mathbb{N}$. I don't understand how I can apply this definition to get a singleton set in $\mathbb{N}$ that's open. Furthermore, I don't even understand what $U$ is in that definition, is it the singleton set or $\mathbb{Q}$ in this context?

Can you explain how to use the above definition to get the result the Napkin did?

Answer 1

Denote by $B(x,r)$ the open ball centered on $x$ with radius $r >0$. $B(x,r)$ is an open subset of the reals... good to know that an open ball is open!

Consider $n \in \mathbb N$. Then $B(n, 1/2) \cap \mathbb N = \{n\}$ and therefore any singleton of $\mathbb N$ is an open subset.

Now if $\mathbb Q$ and $\mathbb N$ would be homeomorphic for the usual topology the image of any open subset of $\mathbb N$ would be isomorphic to its image under the homeomorphism. But that can’t be as it exists open singleton in $\mathbb N$ while any open subset of $\mathbb Q$ contains an infinite number of rationals.

Answer 2

How do open sets in a subspace look like?

If $Y \subset X$ is a subset, then the open sets (of the subspace topology, but you can also think of metric spaces and subspaces) are exactly given by sets of the form $Y \cap U$, where $U$ is an open set in $X$. If you now take a small ball $B_{<\epsilon}(n) = (n-\epsilon,n + \epsilon)$ around a natural number $n$ (small here means that it contains no other natural number), then the intersection with $\mathbb{N}$ only consists of the natural number itself. Thus every singleton $\lbrace n \rbrace \subset \mathbb{N}$ is open and therefore every subset $M \subset \mathbb{N}$ is open as unions of open sets are open.

If we take some open interval around a rational number, we will still find infinitely many rational numbers in that interval, which is why singletons in $\mathbb{Q}$ are not open.

Therefore they cannot be homeomorphic.

Answer 3

The topology of $\mathbb{Q}$ and $\mathbb{N}$ is inherited from $\mathbb{R}$, i.e. a subset $U$ of $\mathbb{Q}$ (or $\mathbb{N}$) is open if and only if there exists an set $U'$ that is open in $\mathbb{R}$ such that $U=U'\cap \mathbb{Q}$ (or $U=U'\cap \mathbb{N}$).

Then the argument is clear. Pick a single element $x\in \mathbb{Q}$. Then you can never find an interval $I$ in $\mathbb{R}$ such that $I \cap \mathbb{Q}=\{x\}$. But for a single element $w\in \mathbb{N}$, the interval $I$ of length say 1/2 centered at $w$ gives us $I\cap \mathbb{N}=\{w\}$. Hence singleton sets are not open in $\mathbb{Q}$ but are open in $\mathbb{N}$.